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Fantom [35]
3 years ago
14

Which sets of measurements could be the side lengths of a triangle? Mark all that apply

Mathematics
1 answer:
wel3 years ago
6 0

its e I think


I hope it helps=)

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Write the set of points greater than or equal to −5−5 but strictly less than 44, excluding 00, as a union of two intervals (if y
Alecsey [184]
What we want to draw is the half closed interval:

                                          I=   [-5, 4) excluding 0,

as the union of 2 different intervals say a and b.

Let a be the interval containing the smallest value.

Then the smallest value of a is -5, and the largest possible value of a is 0, not included, if we want to write a as an interval, without "gaps", or "holes".

So a=[-5, 0)

similarly, b=(0, 4).


Answer: [-5, 4)=[-5, 0)∪(0, 4)

7 0
3 years ago
The heights of Karina, Miya and Clint were 155cm, 158cm, and 161cm respectively. What is the height of Roger if the mean of all
Gnesinka [82]

Answer:

refrase the question

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
The earth has an average diameter of 12.800km. If you walked 24 hours a day, 5 days a week (and if you could walk on water)at a
Dimas [21]
1) Circumference of Earth: Diameter x π → 12,800 x π = 40,212.39 km 

2) time in Hour = distance/speed → distance = 40,212.39 and speed = 2 Mi/h 

1st: convert miles into km : 1 mile = 1.609344 km and 2 miles = 3.218688 km
2nd: convert speed into km/h → speed = 3.218688 km/h
And time in hour = 40,212.39 / 3.218688 → time = 12,493 hours  OR in days:
12,493 / 24 = 521 days (OR 104 weeks, including 2 days of rest)

4 0
3 years ago
The directions says determine weather each rate of change is consistent. If it is, find the rate of change and explain what it r
Whitepunk [10]

3:1 is your answer bruuuuuuhhhh

4 0
2 years ago
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Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len
Rama09 [41]

The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is

\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt

In this case, we have

<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> )   ==>   d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )

<em>y(t)</em> = 5 - 2<em>t</em>   ==>   d<em>y</em>/d<em>t</em> = -2

and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then

\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt

=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt

=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt

=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt

=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}

5 0
2 years ago
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